Make a convenient choice of the azimuthal angle [tex]\phi[/tex], say, 0º. You now have an equation in [tex]\theta[/tex] only, and can see immediately that the polar angle (or co-latitude) can't be 90º, so the intersection circle is not an equator of the sphere. Where must it be and what is the circle's area there?

When one sleep deprived person is trying to converse with another, chaos is bound to ensue. I appreciate the help from both of you. I'll admit though that I'm still at a loss. I know what the normal should be so that the problem works out but I don't know how to get that.

(I will also admit that I took vector calculus 6 years ago and I'm not doing such a great job of reteaching myself).

Sorry, Stokes theorem will give the same answer irrespective of the surface. One problem is that the normal is probably negative. Besides that you've the done it correctly. To see why, rotate the space so that [tex]\nabla\times \vec{F} = -\sqrt{3}\vec{k}[/tex]. Now the circle lies in the xy plane. Parameterize the surface in the new coordinate system: [tex]\vec{\phi} = (rcos\theta,rsin\theta,0)[/tex]. If you carry out the surface integral explicitly, you get the same answer, [tex]\sqrt{3}\pi a^2[/tex].